Find the equivalent capacitance between the points A and B.

4 μF

The correct answer is Option (1) → 4 μF

Equivalent capacitance between A and B:

Top branch: two capacitors in series,

$C_{top}=\frac{3 \times 3}{3+3}=1.5 \,\mu F$

Bottom branch: two capacitors in series,

$C_{bottom}=\frac{3 \times 3}{3+3}=1.5 \,\mu F$

Diagonal capacitor (3 μF) parallel to left capacitor of top branch gives,

$C_{left}=3+3=6 \,\mu F$

So effective top branch,

$C_{top}=\frac{6 \times 3}{6+3}=\frac{18}{9}=2 \,\mu F$

Diagonal capacitor (3 μF) parallel to right capacitor of bottom branch gives,

$C_{right}=3+3=6 \,\mu F$

So effective bottom branch,

$C_{bottom}=\frac{3 \times 6}{3+6}=\frac{18}{9}=2 \,\mu F$

Now these two branches are in parallel,

$C_{eq}=2+2=4 \,\mu F$

Final Answer: $C_{eq}=4 \,\mu F$